Comparing double with literal value in C gives different results on 32 bit machines

Question

Can someone please explain why:

double d = 1.0e+300;
printf("%d\n", d == 1.0e+300);

Prints "1" as expected on a 64-bit machine, but "0" on a 32-bit machine? (I got this using GCC 6.3 on Fedora 25)

To my best knowledge, floating point literals are of type double and there is no type conversion happening.

Update: This only occurs when using the -std=c99 flag.

Answer 1

The C standard allows to silently propagate floating-point constant to long double precision in some expressions (notice: precision, not the type). The corresponding macro is FLT_EVAL_METHOD, defined in <float.h> since C99.

As by C11 (N1570), §5.2.4.2.2, the semantic of value 2 is:

evaluate all operations and constants to the range and precision of the long double type.

From the technical viewpoint, on x86 architecture (32-bit) GCC compiles the given code into FPU instructions using x87 with 80-bit stack registers, while for x86-64 architecture (64-bit) it preffers SSE unit (as scalars within XMM registers).

The current implementation was introduced in GCC 4.5 along with -fexcess-precision=standard option. From the GCC 4.5 release notes:

GCC now supports handling floating-point excess precision arising from use of the x87 floating-point unit in a way that conforms to ISO C99. This is enabled with -fexcess-precision=standard and with standards conformance options such as -std=c99, and may be disabled using -fexcess-precision=fast.